S2T2

Jim Randell 11:26 am on 22 February 2022 Permalink Reply
Tags: by: Dr A Alan Taylor   

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  Jim Randell 11:26 am on 22 February 2022 Permalink | Reply

  The grid is (of course) to be completed using Roman numerals.

  The following Python program runs in 50ms.

  Run: [ @replit ]

  from enigma import (int2roman, roman2int, irange, divisors, is_prime, catch, printf)
  
  # find length k romans
  def romans(ns, k=3):
    for n in ns:
      r = int2roman(n)
      if len(r) == k:
        yield r
  
  # multiples for i across
  n = roman2int('XXXVII')
  m1as = list(romans(irange(n, 5000, step=n)))
  
  # multiples for ii across
  n = roman2int('LXXIII')
  m2as = list(romans(irange(n, 5000, step=n)))
  
  # multiples for ii down
  n = roman2int('VII')
  m2ds = list(romans(irange(n, 5000, step=n)))
  
  # squares
  sqs = list(romans(i * i for i in irange(1, 70)))
  
  # i down (= ADG) is a square
  for (A, D, G) in sqs:
  
    # ii across is a multiple of 73
    for (D_, E, F) in m2as:
      if D_ != D: continue
  
      # i across is a multiple of 37
      for (A_, B, C) in m1as:
        if A_ != A: continue
  
        # ii down is a multiple of 7
        for (B_, E_, H) in m2ds:
          if B_ != B or E_ != E: continue
  
          # iii across is a non-prime factor of i down
          for (G_, H_, I) in romans(d for d in divisors(roman2int(A + D + G)) if not is_prime(d)):
            if G_ != G or H_ != H: continue
  
            # answer (must be valid roman)
            ans = catch(roman2int, C + F + I)
            if ans is None: continue
  
            # output solution
            printf("ans = {ans} [{A} {B} {C} / {D} {E} {F} / {G} {H} {I}]")
  

  Solution: Calpurnia is 19.

  The completed grid looks like this:

  

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  • 

    Jim Randell 1:54 pm on 22 February 2022 Permalink | Reply

    Manually:

    There is only one length 3 multiple of 73 (for II across):

    DXI = 511 = 7×73

    And there are only three length 3 squares (for I down):

    XVI = 16 = 4²
    XXV = 25 = 5²
    MDC = 1600 = 40²
    MMD = 2500 = 50²

    Of these, only MDC interlocks with DXI.

    There are only three length 3 multiple of 37 (for I across):

    CXI = 111 = 3×37
    DLV = 555 = 15×37
    MCX = 1110 = 30×37

    Of these, only MCX interlocks with MDC.

    Which leaves II down = “a multiple of VII” that matches “CX_”, it must be:

    CXL = 140 = 20×7

    And III across = “a non-prime divisor of 1600”, that matches “CL_”, it must be:

    CLX = 160 = 1600 / 10

    The grid is now complete and III down = XIX = 19.

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